proof of Abel’s limit theorem

Without loss of generality we may assume $r=1$ , because otherwise we can set $a^{\prime}_{n}:=a_{r}^{n}$ , so that $\sum a^{\prime}_{n}x^{n}$ has radius $1$ and $\sum a^{\prime}$ is convergent if and only if $\sum a_{n}r^{n}$ is. We now have to show that the function $f(x)$ generated by $\sum a_{n}x^{n}$ (with $r=1$ )is continuous from below at $x=1$ if it is defined there. Let $s:=\sum a_{n}$ . We have to show that

\lim_{x\to 1^{-}}f(x)=s.

If $|x|<1$ we have:

	$\displaystyle s-f(x)$	$\displaystyle=\sum_{n=0}^{\infty}a_{n}-\sum_{n=0}^{\infty}a_{n}x^{n}$
		$\displaystyle=\sum_{n=0}^{\infty}(1-x^{n})a_{n}$
		$\displaystyle=(1-x)\sum_{n=1}^{\infty}(x^{n-1}+x^{n-2}+\dots+x+1)a_{n}$
		$\displaystyle=(1-x)\sum_{n=0}^{\infty}(s-s_{n})x^{n}$

with $s_{n}:=\sum_{i=0}^{n}a_{i}$ . Now, since $s-s_{n}\to 0$ as $n\to\infty$ we can choose an $N$ for every $\varepsilon>0$ such that $|s-s_{n}|<\frac{\varepsilon}{2}$ for all $m>N$ . So for every $0<x<1$ we have:

	$\displaystyle\|s-f(x)\|$	$\displaystyle<(1-x)\sum_{n=0}^{m}\|r_{n}\|x^{n}+\frac{\varepsilon}{2}(1-x)\sum_{% n=m+1}^{\infty}x^{n}$
		$\displaystyle<(1-x)\sum_{n=0}^{m}\|r_{n}\|+\frac{\varepsilon}{2}.$

This is smaller than $\varepsilon$ for all $x<1$ sufficiently close to $1$ , which proves

\lim_{x\to r^{-}}\sum a_{n}x^{n}=\sum a_{n}r^{n}=\sum\lim_{x\to r^{-}}a_{n}x^{% n}.

Title	proof of Abel’s limit theorem
Canonical name	ProofOfAbelsLimitTheorem
Date of creation	2013-03-22 14:09:40
Last modified on	2013-03-22 14:09:40
Owner	mathwizard (128)
Last modified by	mathwizard (128)
Numerical id	5
Author	mathwizard (128)
Entry type	Proof
Classification	msc 40A30
Related topic	ProofOfAbelsConvergenceTheorem